Deriving the Michaelis-Menten Equation

Gale Rhodes
Chemistry Department
University of Southern Maine

Revised 2006/07/14

Memorize this derivation as
soon as your encounter it in your text, and you will be able to read
the remainder of the chapter with far greater understanding. For
other suggestions on how to make your study of biochemistry easier,
see Learning
Strategies.

A simple model of enzyme action:

We would like to know how to recognize an enzyme
that behaves according to this model. One way is to look at the
enzyme's kinetic behavior -- at how substrate concentration affects
its rate. So we want to know what rate law such an enzyme would obey.
If a newly discovered enzyme obeys the rate law derived from this
model, then it's reasonable to assume that the enzyme reacts
according to this model. It's not proof that the model is
correct, but at least it tells us that kinetics does not rule it
out.

Let's derive a rate law from this model.

For this model, let V0 be the initial
velocity of the reaction. Then

V0 = kcat [ES].
(2)

The maximum velocity Vmax occurs when
the enzyme is saturated -- that is, when

all enzyme molecules are tied up with S,
or

[ES] = [E]total
.

So Vmax = kcat
[E]total . (3)

We want to express V0 in terms of
measurable quantities, like [S] and
[E]total , so we can see how to test the mechanism
by experiments in kinetics. So we must replace [ES] in (2)
with measurables.

During the initial phase of the reaction, as long
as the reaction velocity remains constant, the reaction is in a
steady state, with ES being formed and consumed at the same
rate. During this phase, the rate of formation of [ES] equals
its rate of consumption. According to model (1),

Rate of formation of ES =
k1[E][S].

Rate of consumption of ES =
k-1[ES] + kcat [ES].

So in the steady state, k-1[ES]
+ kcat [ES] = k1[E][S].
(4)

Remember that we are trying to solve for
[ES] in terms of measurables, so that we can replace it in
(2). First, collect the kinetic constants in (4):

(k-1 + kcat) [ES] =
k1[E][S],

and (k-1 +
kcat)/k1 = [E][S]/[ES].
(5)

To simplify (5), first group the kinetic constants
by defining them as Km :

Km = (k-1 +
kcat)/k1 (6)

and then express [E] in terms of
[ES] and [E]total:

[E] = [E]total -
[ES] (7)

Substitute (6) and (7) into (5):

Km = ([E]total - [ES])
[S]/[ES] (8)

Solve (8) for [ES]: First multiply both
sides by [ES]:

[ES] Km =
[E]total [S] -
[ES][S]

Then collect terms containing [ES] on the
left:

[ES] Km +
[ES][S] = [E]total
[S]

Factor [ES] from the left-hand
terms:

[ES](Km + [S]) =
[E]total [S]

and finally, divide both sides by (Km +
[S]):

[ES] = [E]total
[S]/(Km + [S]) (9)

Substitute (9) into (2): V0 =
kcat [E]total
[S]/(Km + [S]) (10)

Recalling (3), substitute Vmax into
(10) for kcat [E]total:

V0 = Vmax
[S]/(Km + [S]) (11)

This equation expresses the initial rate of
reaction in terms of a measurable quantity, the initial substrate
concentration. The two kinetic parameters, Vmax and
Km , will be different for every enzyme-substrate
pair.

Equation (11), the Michaelis-Menten equation,
describes the kinetic behavior of an enzyme that acts according to
the simple model (1). Equation (11) is of the form

y = ax/(b + x) (does this look
familiar?)

This is the equation of a rectangular hyperbola,
just like the saturation equation for the binding of dioxygen to
myoglobin.

Equation (11) means that, for an enzyme acting
according to the simple model (1), a plot of V0 versus
[S] will be a rectangular hyperbola. When enzymes exhibit
this kinetic behavior, unless we find other evidence to the contrary,
we assume that they act according to model (1), and call them
Michaelis-Menten enzymes.

QUIZ for USM Students

Quiz at first class on enzyme kinetics: Derive
equation (11) from model (1).